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An electron falls through a distance 1.5 cm in a uniformelectric field of magnitude 2.0xx 10^(4)NC^(-1) .Calculate the time it takes to fall through this distance starting from rest. If the direction of the field is reversed keeping its magnitude uncharged , calculate the time taken by a proton to fall through this distance starting from rest. |
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Answer» Solution :Here distance to be covered s `= 1.5 CM = 1.5 xx 10^(-2)m, ` ELECTRON field ` E= 2.0 xx 10 ^(4)N C^(-1)` , charge on an electron /proton `=+- e= 1.6 xx 10 ^(-19)C ,` initial velocityu=0 , mass of electron m`= 9.1 xx 10 ^(-31) ` kg and mass of proton `m. =1.67 xx 10 ^(27) kg.` Acceleration of electron in ELECTRIC field a= `(F)/( m) = ( eE)/( m) ` So as per relation s=ut `+(1)/(2)at ^(2) ` , the time taken by electron to cover the distance ` "" t= sqrt(( 2s)/( a)) = sqrt(( 2sm)/( eE)) =sqrt(( 2xx( 1.5xx10^(-2))xx (9.1xx10^(-31)))/( (1.6xx 10^(-19) ) xx ( 2.0 xx10 ^(4)) ) ` `"" = 2.9xx 10 ^(-9) s ` Similarly time taken by proton t. =`sqrt(( 2sm.)/( eE) ) = sqrt(( 2xx( 1.5xx10^(-2)) xx (1.67xx10 ^(-27)))/( (1.6xx 10^(-19) ) xx ( 2.0 xx10 ^(4))) ` ` ""= 1.3 xx10 ^(-7) s`  `)
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